Solve y=Acosx+Bsinx

$$y = A \cos x + B \sin x$$

Solve for A

$$\left\{\begin{matrix}A=-\frac{B\sin(x)-y}{\cos(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\A\in \mathrm{R}\text{, }&y=B\sin(x)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$A\cos(x)+B\sin(x)=y$$

Subtract B\sin(x) from both sides.

$$A\cos(x)=y-B\sin(x)$$

The equation is in standard form.

$$\cos(x)A=-B\sin(x)+y$$

Divide both sides by \cos(x).

$$\frac{\cos(x)A}{\cos(x)}=\frac{-B\sin(x)+y}{\cos(x)}$$

Dividing by \cos(x) undoes the multiplication by \cos(x).

$$A=\frac{-B\sin(x)+y}{\cos(x)}$$

Solve for B

$$\left\{\begin{matrix}B=-\frac{A\cos(x)-y}{\sin(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\\B\in \mathrm{R}\text{, }&y=A\cos(x)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$A\cos(x)+B\sin(x)=y$$

Subtract A\cos(x) from both sides.

$$B\sin(x)=y-A\cos(x)$$

The equation is in standard form.

$$\sin(x)B=-A\cos(x)+y$$

Divide both sides by \sin(x).

$$\frac{\sin(x)B}{\sin(x)}=\frac{-A\cos(x)+y}{\sin(x)}$$

Dividing by \sin(x) undoes the multiplication by \sin(x).

$$B=\frac{-A\cos(x)+y}{\sin(x)}$$